Constant Sum Group Flows of Graphs

Abstract

As an analogous concept of a nowhere-zero flow for directed graphs, zero-sum flows and constant-sum flows are defined and studied in the literature. For an undirected graph, a zero-sum flow (constant-sum flow resp.) is an assignment of nonzero integers to the edges such that the sum of the values of all edges incident with each vertex is zero (constant h resp.), and we call it a zero-sum k-flow (h-sum k-flow resp.) if the values of the edges are less than k. We extend these concepts to general constant-sum A-flow, where A is an Abelian group, and consider the case $A={\mathbb{Z}}_k$ the additive Abelian cyclic group of integer congruences modulo k with identity 0. In the literature, a graph is alternatively called ${\mathbb{Z}}_k$-magic if it admits a constant-sum ${\mathbb{Z}}_k$-flow, where the constant sum is called a magic sum or an index for short. We define the set of all possible magic sums such that G admits a constant-sum ${\mathbb{Z}}_k$-flow to be $I_k\left(G\right)$ and call it the magic sum spectrum, or for short, the index set of G with respect to ${\mathbb{Z}}_k$. In this article, we study the general properties of the magic sum spectrum of graphs. We determine the magic sum spectrum of complete bipartite graphs $K_{m,n}$ for $m\ge n\ge 2$ as the additive cyclic subgroups of ${\mathbb{Z}}_k$ generated by $\frac{k}{d}$, where $d=gcd(m-n,k)$. Also, we show that every regular graph G with a perfect matching has a full magic sum spectrum, namely, $I_k\left(G\right)={\mathbb{Z}}_k$ for all $k\ge 3$. We characterize a 3-regular graph so that it admits a perfect matching if and only if it has a full magic sum spectrum, while an example is given for a 3-regular graph without a perfect matching which has no full magic sum spectrum. Another example is given for a 5-regular graph without a perfect matching, which, however, has a full magic sum spectrum. In particular, we completely determine the magic sum spectra for all regular graphs of even degree. As a byproduct, we verify a conjecture raised by Akbari et al., which claims that every connected $4k$-regular graph of even order admits a 1-sum 4-flow. More open problems are included.

1. Introduction and Background

In this article, only finite undirected graphs are considered. A graph G has the edge set $E\left(G\right)$ and vertex set $V\left(G\right)$. For any additive Abelian group A, let $A^{*}=A-\left\{0\right\}$ be the set of all nonzero elements and 0 the additive identity of A. Given a graph G, a mapping $f:E\left(G\right)\to A^{*}$ is also called an edge labeling of G. A graph G admits a constant-sum A-flow or is said to be A-magic if there exists an edge labeling such that the induced vertex labeling $f^{+}:V\left(G\right)\to A$ defined by $f^{+}\left(v\right)={\sum }_{uv\in E\left(G\right)}f\left(uv\right)$ is a constant map. We call the constant a magic sum of G with respect to A, an index for short. We then denote the set of all constants r such that the graph G admits a constant-sum A-flow with an index r by $I_A\left(G\right)$ and call it the magic sum spectrum, or the index set for short, of G with respect to A.

In general, a graph may admit more than one edge labeling to be an A-magic graph. No generally efficient algorithm is known for finding magic labeling for general graphs. A. Kotzig and A. Rosa used the same term in [1], but their notion of magic labeling is different from what we consider here. It is well known [2,3,4] that a graph G is $\mathbb{N}$-magic if and only if every edge of G is contained in a $(1,2)$-factor and every pair of edges is separated by this $(1,2)$-factor. For a list of properties of $\mathbb{N}$-magic graphs, see [5,6,7]. Richard Stanley first studied $\mathbb{Z}$-magic graphs in [8,9], and he demonstrated that the study of magic labelings can be reduced to solving a system of linear diophantine equations. More references can be seen in the survey article by Gallian [10].

Note that one special case of magic sums of A-magic graphs while $A=\mathbb{Z}$ with magic sum constant zero has been studied intensively, namely zero-sum flows, which was initiated by S. Akbari et al. [11] in 2009. For an undirected graph G, a zero-sum flow is an assignment of nonzero integers to the edges such that the sum of the values of all edges incident with each vertex is zero. A zero-sum k-flow is a zero-sum flow whose values are integers with an absolute value less than k. One may treat zero-sum flows as an undirected analog of nowhere-zero flows for directed graphs. A celebrated conjecture of Tutte in 1954 says that every bridgeless graph has a nowhere-zero 5-flow. There is a more general concept of a nowhere-zero flow that uses bidirected edges instead of directed ones, first systematically developed by Bouchet [12] in 1983. Bouchet raised the conjecture that every bidirected graph with a nowhere-zero integer flow has a nowhere-zero 6-flow, which is still unsettled. In 2010 S. Akbari et al. raised a conjecture (called Zero-Sum 6-Flow Conjecture) for zero-sum flows similar to Tutte’s 5-flow conjecture for nowhere-zero flows as follows: if G is a graph with a zero-sum flow, then G admits a zero-sum 6-flow. It was proved by Akbari et al. [13] that the above Zero-Sum 6-Flow Conjecture is equivalent to Bouchet’s 6-Flow Conjecture for bidirected graphs. The study of zero-sum flows was extended to constant-sum flows by T.-M. Wang et al. [14] in 2011 and also studied by Akbari et al. [15,16,17,18] as 0-sum and 1-sum flows with related topics.

The index sets $I_A\left(G\right)$ of various graph classes G were studied and calculated. For example, let $A=\mathbb{Z}$ and write the magic sum index set as $I\left(G\right)$. In [14], in 2011, T.-M. Wang et al. obtained the following result regarding the index sets of r-regular graphs (except the case $r=4k,k\ge 1) of order n$:

$$I\left(G\right)=\left\{\begin{array}{c}{\mathbb{Z}}^{*}=\mathbb{Z}-\left\{\mathbf{0}\right\},r=1.\\ \mathbb{Z},r=2and G contains even cycles only.\\ \mathbf{2}{\mathbb{Z}}^{*},r=2and G contains an odd cycle.\\ \mathbf{2}\mathbb{Z},r\ge 3,reven and n odd.\\ \mathbb{Z},r\ge 3,r\ne 4k,and n even.\end{array}\right.$$

In this article, graphs admitting constant-sum ${\mathbb{Z}}_k$-flows are studied, i.e., $A={\mathbb{Z}}_k$. We refer to $I_k\left(G\right)$ as the magic sum index set with respect to ${\mathbb{Z}}_k$, for $k\ge 2$. Note that the case of constant-sum ${\mathbb{Z}}_2$-flows is not hard to calculate since every edge of a graph with ${\mathbb{Z}}_2$-flows must be labeled with 1, and hence, the magic sum index set is completely determined by the degree sequence of the graph. Therefore, the magic sum spectra $I_k\left(G\right)$ are of interest for $k\ge 3$. In this paper, we completely determine the magic sum index sets of all regular graphs of even degree and characterize 3-regular graphs with a full magic constant spectrum $I_k\left(G\right)={\mathbb{Z}}_k$ for all $k\ge 3$. We also give certain examples for index sets in other cases.

Below, we list the major results of the magic sum spectrum $I_k\left(G\right)$ when $A={\mathbb{Z}}_k$, among others, and prove them in the subsequent sections.

Theorem 1.

For $m\ge n\ge 2$ and $k\ge 3$, the magic sum spectrum of the complete bipartite graph $K_{m,n}$ is $I_k\left(K_{m,n}\right)=\langle \frac{k}{d}\rangle$, where $d=gcd(m-n,k)$, and $\langle a\rangle$ denotes the additive subgroup of ${\mathbb{Z}}_k$ generated by the element a. Or, equivalently,

$$I_k\left(K_{m,n}\right)=\left\{\begin{array}{c}\left\{0\right\},ifgcd(m-n,k)=1.\\ {\mathbb{Z}}_d\cong \langle \frac{k}{d}\rangle ,ifgcd(m-n,k)=d,1<d\le k.\end{array}\right.$$

Theorem 2.

Let G be an r-regular graph ($r\ge 2$) which admits a perfect matching. Then, $I_k\left(G\right)={\mathbb{Z}}_k$ for all $k\ge 3$.

Theorem 3.

Let G be a 3-regular graph. Then, $I_k\left(G\right)={\mathbb{Z}}_k$ for all $k\ge 3$ if and only if G admits a perfect matching.

Theorem 4.

There exists an example of a 3-regular graph $G_3$ containing no perfect matching such that

$$I_k\left(G_3\right)=\left\{\begin{array}{c}\left\{0\right\},k=3,\\ {\mathbb{Z}}_4\setminus \left\{0\right\},k=4,\\ {\mathbb{Z}}_k,k\ge 5.\end{array}\right.$$

Theorem 5.

There exists an example of a 5-regular graph $G_5$, without any perfect matching, such that $I_k\left(G_5\right)={\mathbb{Z}}_k$ for all $k\ge 3$.

Theorem 6.

If G is an even–regular graph with an odd number of vertices, then $I_k\left(G\right)=2{\mathbb{Z}}_k$ for all $k\ge 3$.

Theorem 7.

If G is a connected regular graph of degree $2m$, with $m>1$ odd and an even number of vertices, then $I_k\left(G\right)={\mathbb{Z}}_k$ for all $k\ge 3$.

Theorem 8.

If G is a connected regular graph of degree $2m$, with m even and an even number of vertices, then $I_k\left(G\right)={\mathbb{Z}}_k$ for $k\ge 3$.

Inspired by extensive studies of magic labelings in the literature with infinite cyclic Abelian group $\mathbb{Z}$, one naturally considers the finite cyclic group ${\mathbb{Z}}_k$ of congruence modulo k. Therefore, in this article, we study magic labelings with an emphasis on the group ${\mathbb{Z}}_k$, which is closely related to the infinite cyclic case $\mathbb{Z}$. In the concluding section, we mention the observation that this work can be further extended to more general cases via the fundamental theorem of finite Abelian groups and other structure theorems of Abelian groups. For more group theoretical discussions, please refer to the book by Rotman [19], for instance.

2. Magic Sum Spectra for General Graphs

In this section, we investigate the properties of magic sum spectra for general graphs. For the index sets of the vertex disjoint union of graphs, we have the following formula.

Proposition 1.

Let $H_1$ and $H_2$ be vertex disjoint graphs. Then, $I_k(H_1\cup H_2)=I_k\left(H_1\right)\cap I_k\left(H_2\right)$. In particular, let $nG$ be the vertex disjoint union of n copies of G, then $nG$ has the same index set as G, that is, $I_k\left(nG\right)=I_k\left(G\right)$.

Proof. 

On one hand, if $r\in I_k(H_1\cup H_2)$, then $r\in I_k\left(H_1\right)$ and $r\in I_k\left(H_2\right)$ simply consider the restriction of the labeling of $H_1\cup H_2$ over $H_1$ and $H_2$, respectively. On the other hand, if $r\in I_k\left(H_1\right)$ and $r\in I_k\left(H_2\right)$, then $r\in I_k(H_1\cup H_2)$, since $H_1$ and $H_2$ are vertex disjoint, hence edge-disjoint. □

We have the following basic observations over the indices of an edge-disjoint union of ${\mathbb{Z}}_k$-magic graphs. Let, in particular, $G=G_1\oplus G_2\oplus \cdots \oplus G_m$ be the edge-disjoint union of spanning subgraphs $G_1,G_2,\cdots$ and $G_m$, that is, G admits a factorization with factors $G_1,G_2,\cdots ,G_m$. Suppose for fixed k the graphs $G_i$ is ${\mathbb{Z}}_k$-magic with index $r_i$ for $i=1,\cdots ,k$. Then, we have that $G=G_1\oplus G_2\oplus \cdots \oplus G_m$ is ${\mathbb{Z}}_k$-magic with an index ${\sum }_{i=1}^m{r}_i$. The reason is that at each vertex v of $G_1\oplus G_2\oplus \cdots \oplus G_m$, the incident edges consist of the incident edges of v in $G_1,G_2,\cdots ,G_m$. We organize these facts as follows.

Proposition 2.

Let $G=G_1\oplus G_2\oplus \cdots \oplus G_m$ be the factorization of spanning subgraphs $G_1,G_2,\cdots$ and $G_m$. Suppose for fixed k the graph $G_i$ is ${\mathbb{Z}}_k$-magic with index $r_i$ for $i=1,\cdots ,k$. Then, we have the following:

(1) 

If $r_1=r_2=\cdots =r_m=0$, then $G_1\oplus G_2\oplus \cdots \oplus G_m$ has an index of 0.

(2) 

$G=G_1\oplus G_2\oplus \cdots \oplus G_m$ is ${\mathbb{Z}}_k$-magic with an index of ${\sum }_{i=1}^m{r}_i$.

(3) 

$I_k\left(G_1\right)+I_k\left(G_2\right)+\cdots +I_k\left(G_m\right)\subseteq I_k(G_1\oplus G_2\oplus \cdots \oplus G_m)$, and in particular, $I_k(G_1\oplus G_2\oplus \cdots \oplus G_m)={\mathbb{Z}}_k$ if $I_k\left(G_i\right)={\mathbb{Z}}_k$ for some i.

Remark 1.

Note that in the previous Proposition, $I_k\left(G_1\right)+I_k\left(G_2\right)+\cdots +I_k\left(G_m\right)\cancel{⫆}{I}_k(G_1\oplus G_2\oplus \cdots \oplus G_m)$ in general. That is, not necessarily every index of $G_1\oplus G_2\oplus \cdots \oplus G_m$ comes from the sum of indices of $G_1,G_2,\cdots$ and $G_m$.

Remark 2.

In case the graphs $G_1,G_2,\cdots$ and $G_m$ are ${\mathbb{Z}}_k$-magic with indices 0 for each $G_i$, where $i=1,\cdots ,k$, then we have the resulting arbitrary union graph ${\bigcup }_{i=1}^k{G}_i$, which is formed by attaching these m graphs in any way, and it is still ${\mathbb{Z}}_k$-magic with an index of 0.

One direct application of the above observation is that the Cartesian product of ${\mathbb{Z}}_k$-magic graphs is still ${\mathbb{Z}}_k$-magic with an index of the sum of corresponding indices:

Proposition 3.

For fixed k, let the graphs $H_i$ be ${\mathbb{Z}}_k$-magic with index $r_i$ for $i=1,\cdots ,s$. Then, the Cartesian product $H_1\times H_2\times \cdots \times H_s$ has an index of ${\sum }_{i=1}^s{r}_i$, and $I_k\left(H_1\right)+I_k\left(H_2\right)+\cdots +I_k\left(H_s\right)\subseteq I_k(H_1\times H_2\times \cdots \times H_s)$. In particular, the index set $I_k(H_1\times H_2\times \cdots \times H_s)={\mathbb{Z}}_k$ whenever $I_k\left(H_i\right)={\mathbb{Z}}_k$ for some $1\le i\le s$.

Proof. 

It suffices to show that the Cartesian product of two ${\mathbb{Z}}_k$-magic graphs is ${\mathbb{Z}}_k$-magic with an index that is the sum of the original indices. Let $|V\left(H_1\right)|=m_1$ and $|V\left(H_2\right)|=m_2$. Decompose $H_1\times H_2$ as graphs $K_1$ and $K_2$ such that $V\left(K_1\right)=V\left(K_2\right)=V(H_1\times H_2)$, where $E\left(K_1\right)=\left\{(u_i,v)(u_j,v)\right|u_i{u}_j\in E\left(H_1\right)\}$ and $E\left(K_2\right)=\left\{(u,v_i)(u,v_j)\right|v_i{v}_j\in E\left(H_2\right)\}$. Then, $K_1\cong m_2{H}_1$ and $K_2\cong m_1{H}_2$, thus having indices $r_1$ and $r_2$, respectively. Then, from the previous facts, $H_1\times H_2=K_1\oplus K_2$ has an index $r_1+r_2$, and $I_k\left(H_1\right)+I_k\left(H_2\right)\subseteq I_k(H_1\times H_2)$. Hence, the proof is complete. □

However, in general, the index set will not be the full ${\mathbb{Z}}_k$. Note that the index set in general is a subset of ${\mathbb{Z}}_k$, not necessarily a subgroup. Note that if $r\in I_k\left(G\right)$, then $-r\in I_k\left(G\right)$ simply by reversing the sign of the labels on each edge. If one can show that for any $r,r^{\prime }\in I_k\left(G\right)$ we have $r+r^{\prime }\in I_k\left(G\right)$, then $I_k\left(G\right)$ is a subgroup of ${\mathbb{Z}}_k$. We have the following necessary condition for a ${\mathbb{Z}}_k$-magic graph with magic sum r.

Proposition 4.

For any ${\mathbb{Z}}_k$-magic labeling f of a graph G with index r, we have

$$2\sum _{e\in E\left(G\right)}f\left(e\right)\equiv r·\left|V\left(G\right)\right|\left(modk\right)$$

Proof. 

Summing the vertex sums will count each edge label twice. □

However, the converse of Proposition 4 is not true in general. By the above necessary condition, for a graph G with $V\left(G\right)$ odd and for even k, it implies that r must be an even number, as a representative in the congruence class modulo k. Therefore, we have the following facts.

Corollary 1.

For a graph G with odd vertices and for even k, we have that $I_k\left(G\right)⫅2{\mathbb{Z}}_k⫋{\mathbb{Z}}_k$.

We have the following examples of the index sets of cycles (connected 2-regular graphs) and the Cartesian product of cycles.

Proposition 5.

Let $C_n$ be an n-cycle, where $n\ge 3$, and k be a positive integer. We have the following:

(1) 

$I_k\left(C_n\right)=2{\mathbb{Z}}_k^{*}=\{2x:x\ne 0,x\in {\mathbb{Z}}_k\}$, for n odd.

(2) 

$I_k\left(C_n\right)={\mathbb{Z}}_k$, for n even.

Proof. 

Note that in any ${\mathbb{Z}}_k$-magic labeling of a cycle, the edges should alternatively be labeled the same. Therefore, for n odd, the labels on all edges are all the same. Therefore, $I_k\left(C_n\right)=2{\mathbb{Z}}_k^{*}=\{2x:x\ne 0,x\in {\mathbb{Z}}_k\}$ for n odd.

On the other hand, for n even, we label the edges $1,x-1,1,x-1,\cdots ,1,x-1$ for $x\in {\mathbb{Z}}_k\setminus \left\{1\right\}$ to obtain the index x, and $2,-1,2,-1,\cdots ,2,-1$ to obtain the index 1. Therefore, $I_k\left(C_n\right)={\mathbb{Z}}_k$ for n even. □

Remark 3.

Note that $2{\mathbb{Z}}_k^{*}={\mathbb{Z}}_k^{*}$ for k is odd, and $2{\mathbb{Z}}_k^{*}=\{0,2,\cdots ,k-2\}=2{\mathbb{Z}}_{\frac{k}{2}}$ for k is even. One may see from the above proposition that $0\in I_k\left(C_n\right)$ if and only if k is even for n odd.

As corollaries of the above observations, we have the following example for the index set of the hypercube graphs $Q_n$:

Proposition 6.

For fixed k and $n\ge 2$, the index set of the hypercube is $I_k\left(Q_n\right)={\mathbb{Z}}_k$. Note that $Q_n=\underset{ncopies}{\underbrace{P_2\times P_2\times \cdots \times P_2}}$.

Proof. 

Note that $I_k\left(Q_2\right)=I_k\left(C_4\right)={\mathbb{Z}}_k$ by Proposition 5. Then, by Proposition 2, we have $I_k\left(Q_n\right)={\mathbb{Z}}_k$. □

Moreover, we have the following examples for the index sets of toroidal grids $C_{m_1}\times C_{m_2}\times \cdots \times C_{m_t}$:

Proposition 7.

For fixed k and $n\ge 2$, the index set of a toroidal grid $G=C_{m_1}\times C_{m_2}\times \cdots \times C_{m_t}$ is

$$I_k\left(G\right)=\left\{\begin{array}{c}{\mathbb{Z}}_k,if some m_{i}is even,\\ 2{\mathbb{Z}}_k,otherwise.\end{array}\right.$$

Proof. 

Directly from the Proposition 3 and the Proposition 5. □

In general, if G is ${\mathbb{Z}}_k$-magic, then it is ${\mathbb{Z}}_{mk}$-magic for k and m positive integers. Therefore, it suffices to consider the cases of prime numbers when one studies the ${\mathbb{Z}}_k$-magicness of a graph. We have the following relationship of index sets with respect to k and $mk$:

Proposition 8.

Let $m,k$ be positive integers. Then, $m·I_k\left(G\right)\subseteq I_{mk}\left(G\right)$ and $|I_k\left(G\right)|\le |I_{mk}\left(G\right)|$.

Proof. 

Note that if G is ${\mathbb{Z}}_k$-magic with index $r\left(modk\right)$, then it is ${\mathbb{Z}}_{mk}$-magic with index $mr\left(modmk\right)$, since we may simply use the ${\mathbb{Z}}_k$-magic labeling with index r, multiplying on each edge label by m, which turns out to be a ${\mathbb{Z}}_{mk}$-magic labeling with index $mr$. On the other hand, $m{r}_i\equiv m{r}_j\left(modmk\right)$ implies $r_i\equiv r_j\left(modk\right)$, thus $|I_k\left(G\right)|\le |I_{mk}\left(G\right)|$. □

Remark 4.

From the above observation, one may see that for checking the ${\mathbb{Z}}_k$-magicness of a graph, it suffices to look at the prime numbers k, instead of all positive integers k.

Another useful fact is as follows:

Lemma 1.

Any Eulerian graph with an even size has a zero index.

Proof. 

Label 1 and $-1$ alternately on a Eulerian cycle. □

3. Magic Sum Spectra of Complete Bipartite Graphs

We determine the index sets of complete bipartite graphs $K_{m,n}$ for $m,n\ge 1$ in this section. Note that these examples are not regular graphs in most cases. One can see that the index x of a constant-sum ${\mathbb{Z}}_k$-flow for $K_{m,n}$ is the solution of the linear congruence equation $(m-n)x\equiv 0\left(modk\right)$, since $mx$ and $nx$ are both the total sum of all edge ${\mathbb{Z}}_k$-labels over the graph $K_{m,n}$, as seen by adding up m of the same indices x on one partite set or n of the same indices x on the other partite set over the graph $K_{m,n}$.

If $m=1$ or $n=1$, $K_{m,n}$ is the star graph, then by the solutions to the above linear congruence equation, we have $I_k\left(K_{1,1}\right)={\mathbb{Z}}_k^{*}$, and $I_k\left(K_{1,n}\right)=\{\frac{k}{d},\frac{2k}{d},\cdots ,\frac{(d-1)k}{d}\}$ for $n\ge 2$ and $d=gcd(n-1,k)>1$. Note that $I_k\left(K_{1,n}\right)=\varnothing$ for $n\ge 2$ and $gcd(n-1,k)=1$, and in particular, $0\notin I_k\left(K_{1,n}\right)$ in any case.

Now, we study the case for $K_{m,n}$ for $m,n\ge 2$ in the following.

Proposition 9.

$0\in I_k\left(K_{m,n}\right)$ for $m,n\ge 2$ and all $k\ge 3$.

Proof. 

We have the following two cases:

Case 1. m or n is even. Say m is even, we decompose $K_{m,n}$ into an edge-disjoint union of copies of $K_{2,n}$. Then, we label the edges of $K_{2,n}$ by the following rules. Labeling 1 and $-1$ on each pair of edges incident to the first $n-1$ degree 2 vertices, then labeling, if necessary (whenever n is odd), 2 and $-2$ on the last pair of edges. Therefore, we obtain an index 0 over each copy of $K_{2,n}$ and hence an index 0 over the whole $K_{m,n}$ for when m or n is even.

Case 2. m and n are both odd. Note that $K_{3,3}$ can be labeled with $\pm 1$ and $\pm 2$ to obtain the index 0. In this case, consider $K_{m,n}$ as the edge-disjoint sum of $K_{3,3}$ and $K_{m-3,n-3}$. The part $K_{m-3,n-3}$ would have an index of 0 by Case 1, together with $K_{3,3}$, it gives rise to an index of 0 .

□

Now, we are in a position to obtain the index set of the complete bipartite graph $K_{m,n}$ and resolve Theorem 1:

Proof. 

Note that the index x of a constant-sum ${\mathbb{Z}}_k$-flow for $K_{m,n}$ is the solution of the linear congruence equation $(m-n)x\equiv 0\left(modk\right)$.

In the case that $gcd(m-n,k)=1$, then $(m-n)x\equiv 0\left(modk\right)$ has the unique solution 0, hence $I_k\left(K_{m,n}\right)=\left\{0\right\}$ since $0\in I_k\left(K_{m,n}\right)$, as shown in Proposition 9.

In the case that $gcd(m-n,k)=d>1$, then there are d solutions $\left(modk\right)$ for $(m-n)x\equiv 0\left(modk\right)$, namely, $0,\frac{k}{d},\frac{2k}{d},\cdots ,\frac{(d-1)k}{d}$. Note that $\{0,\frac{k}{d},\frac{2k}{d},\cdots ,\frac{(d-1)k}{d}\}=\langle \frac{k}{d}\rangle \cong {\mathbb{Z}}_d$. Then, we may obtain all other possible indexes $x\in {\mathbb{Z}}_d^{*}\cong \{\frac{k}{d},\frac{2k}{d},\cdots ,\frac{(d-1)k}{d}\}$ by the following observations. Without loss of generality, we may assume that $m\ge n\ge 2$. Decompose the graph $K_{m,n}$ into two parts: one is regular bipartite $K_{n,n}$ and the other is $K_{m-n,n}$. Note that we write the part $K_{m-n,n}$ as a bipartite graph with bipartition $(A,B)$, as in Figure 1.

We claim that there exists appropriate edge labeling on the part $K_{m-n,n}$ such that the vertex sums over the partite set A of $m-n$ vertices are $x\in {\mathbb{Z}}_d^{*}$ and the vertex sums over the partite set B of n vertices are 0; then, since $I_k\left(K_{n,n}\right)={\mathbb{Z}}_k$ for all $k\ge 3$ by the Theorem 10, the desired labeling is obtained by combining the labels from two parts. To prove this claim, we have two cases, as follows.

When n is odd, note that the degrees of vertices in the partite set B are $m-n$. We label the edges of $K_{m-n,n}$ in an ordering of vertices in B by $\underset{m-n}{\underbrace{x,\cdots ,x}}$, $\underset{m-n}{\underbrace{-x,\cdots ,-x}}$, $\underset{m-n}{\underbrace{x,\cdots ,x}}$, $\underset{m-n}{\underbrace{-x,\cdots ,-x}}$, ⋯, until $\underset{m-n}{\underbrace{x,\cdots ,x}}$. Then, the partial vertex sums over the vertices in B with respect to A are either $(m-n)·x$ or $(m-n)·(-x)$, and both $\equiv 0\left(modk\right)$ since x is a solution to $(m-n)x\equiv 0\left(modk\right)$. On the other hand, the vertex sums over the vertices in A are exactly x.

On the other hand, we consider the case when n is even. If 2$x\ne 0\left(modk\right)$, we label the edges as in the previous case by $\underset{m-n}{\underbrace{x,\cdots ,x}}$, $\underset{m-n}{\underbrace{-x,\cdots ,-x}}$, $\underset{m-n}{\underbrace{x,\cdots ,x}}$, $\underset{m-n}{\underbrace{-x,\cdots ,-x}}$, ⋯, except the last two groups we use $\underset{m-n}{\underbrace{2x,\cdots ,2x}}$ and $\underset{m-n}{\underbrace{-x,\cdots ,-x}}$ instead. Then, the new labeling does the job.

In case $2x=0\left(modk\right)$, we see in this case, k is even and $x=\frac{k}{2}\ne 0$. We claim that $m-n$ must be even. Note that $x=\frac{k}{2}\ne 0$ is an element of order 2 in the group ${\mathbb{Z}}_k$ and hence also an element of order 2 in the group ${\mathbb{Z}}_d\cong \langle \frac{k}{d}\rangle$ since $d|k$. Therefore, $2|d=gcd(m-n,k)$, thus $m-n$ must be even.

Then, in the following, we may give a new labeling under the situation that n is even, $m-n$ is even, and k is even. In the partite set A, we pair off the vertices as $\{u_1,u_2\},\{u_3,u_4\}\cdots ,\{u_{m-n-1},u_{m-n}\}$ and label at each 4-cycle containing these two-vertex sets $\{u_1,u_2\},\{u_3,u_4\}\cdots ,\{u_{m-n-1},u_{m-n}\}$, respectively, by $\frac{k}{2}-1,1,-1,1-\frac{k}{2}$ consecutively, as in Figure 2.

Then, it is a straightforward computation to figure out that the partial vertex sums over $u_1,u_2,u_3,u_4,\cdots \cdots ,u_{m-n-1},u_{m-n}$ on these 4-cycles are all $\frac{k}{2}\equiv \frac{-k}{2}\left(modk\right)$, and the partial vertex sums over vertices related in the other partite set B are all zero.

Lastly, note that the remaining edges not used in the above $C_4$ constructions induce an even subgraph (in which every degree is even) of $K_{m-n,n}$, which is Eulerian of even size in each component. Then, as in Lemma 1, we label 1 and $-1$ consecutively on the edges of a Eulerian cycle, which would give partial vertex sums of zero over the even subgraph. Combining all the above partial sums together, we obtain a new labeling as desired on the part $K_{m-n,n}$. □

4. Magic Sum Spectra for Regular Graphs

We study magic sum spectra for regular graphs in this section and first start with several lemmas.

Lemma 2.

For any cubic graph G with a 1-factor, $I_k\left(G\right)={\mathbb{Z}}_k$ for all $k\ge 3$.

Proof. 

Assume $k\ge 3$. Let M be the 1-factor of G. Since $I_k\left(M\right)={\mathbb{Z}}_k^{*}$ and $G\setminus M$ is a 2-regular graph with index 2 via labeling 1 on all edges, then $2+{\mathbb{Z}}_k^{*}={\mathbb{Z}}_k\setminus \left\{2\right\}\subset I_k\left(G\right)$. It remains to show $2\in I_k\left(G\right)$. Labeling $-2$ on all edges of M and 2 on all edges of $G\setminus M$, it is easy to see that 2 is an index. □

Lemma 3.

Let G be a regular graph. Suppose G has a factorization containing a 1-factor $M_1$ and a 2-factor $M_2$, then $I_k\left(G\right)={\mathbb{Z}}_k$ for all $k\ge 3$.

Proof. 

Assume that $k\ge 3$. Note that G may be factored as an edge-disjoint sum of two ${\mathbb{Z}}_k$-magic spanning subgraphs $G=(M_1\oplus M_2)\oplus (G\setminus (M_1\oplus M_2)$, where $M_1$ is a 1-factor and $M_2$ is a 2-factor. Therefore, $I_k(M_1\oplus M_2)={\mathbb{Z}}_k$ by Lemma 2, and this lemma follows by $I_k\left(G\right)=I_k((M_1\oplus M_2)\oplus (G\setminus (M_1\oplus M_2))\subseteq I_k\left((M_1\oplus M_2)\right)+I_k((G\setminus (M_1\oplus M_2))={\mathbb{Z}}_k$. □

We are in a position to show a sufficient condition to have the full magic sum spectrum ${\mathbb{Z}}_k$ for general r-regular graphs. It is well known one has the following theorem for even–regular graphs:

Theorem 9

(Petersen, 1891 [20]). A nonempty graph G is $2m$-regular for some $m\ge 1$ if and only if G admits 2-factorization.

Theorem 10.

Let G be a r-regular graph ($r\ge 2$) which admits a 1-factor, then $I_k\left(G\right)={\mathbb{Z}}_k$ for all $k\ge 3$.

Proof. 

Assume that $k\ge 3$. Let M be the 1-factor. Note that $G=M\oplus (G\setminus M)$ and $G\setminus M$ is an $(r-1)$-regular graph. Since for the 1-factor $I_k\left(M\right)={\mathbb{Z}}_k^{*}$, and $G\setminus M$ has indices $r-1$ and $1-r$ by labeling 1 and $-1$, respectively, on edges, we have that $I_k\left(G\right)=I_k(M\oplus (G\setminus M))\supseteq (r-1)+{\mathbb{Z}}_k^{*}$ and $I_k\left(G\right)=I_k(M\oplus (G\setminus M))\supseteq (1-r)+{\mathbb{Z}}_k^{*}$. If $(r-1)+{\mathbb{Z}}_k^{*}\ne (1-r)+{\mathbb{Z}}_k^{*}$, then we are done with $I_k\left(G\right)={\mathbb{Z}}_k$, since $(r-1)+{\mathbb{Z}}_k^{*}={\mathbb{Z}}_k^{*}\setminus \{r-1\}$ and $(1-r)+{\mathbb{Z}}_k^{*}={\mathbb{Z}}_k^{*}\setminus \{1-r\}$. In case $(r-1)+{\mathbb{Z}}_k^{*}=(1-r)+{\mathbb{Z}}_k^{*}$, which implies $r-1\equiv 1-r$, that is, $2(r-1)\equiv 0\left(modk\right)$. Therefore, we have that $G\setminus M$ has an index of 0 by labeling 2 on the edges. Hence, $I_k\left(G\right)=I_k(M\oplus (G\setminus M))\supseteq 0+{\mathbb{Z}}_k^{*}={\mathbb{Z}}_k^{*}$ in this case, and it remains to show that $0\in I_k\left(G\right)$. For the case where r is even, note that G has an even order, hence G is a Eulerian graph of even size and $0\in I_k\left(G\right)$ by Lemma 1. For the case where r is odd, since $G\setminus M$ is an even–regular graph since by Petersen’s Theorem 9 it has a 2-factorization, then G has a 1-factor and a 2-factor by Lemma 3, and $I_k\left(G\right)={\mathbb{Z}}_k,k>2$. Then, we are done with the proof. □

Remark 5.

Conversely, however, for an r-regular graph that has a full magic sum spectrum, it may or may not admit a perfect matching except for 3-regular graphs. We show this fact later. Note that there are examples for regular graphs without 1-factor for which the index set is not full ${\mathbb{Z}}_k,k\ge 3$.

5. A 3-Regular Graph without Full Magic Sum Spectrum

Let $G_3$ be the example made by Petersen for a 3-regular graph containing no perfect matchings (see Figure 3). We show that the 3-regular graph $G_3$ has no full magic constant spectrum.

Lemma 4.

$\pm 1\notin I_3\left(G_3\right)$.

Proof. 

In Figure 3, we have $b+c=d+e,c+d=e+f$, which implies $b+f=2d$. Suppose $1\in I_3\left(G_3\right)$. Without loss of generality, there exists one of three edge labels of the top claw that can be 1, say $a=1$. Therefore, this implies $0=b+f=2d$ in ${\mathbb{Z}}_3$; the nonzero solution for d does not exist, which is a contradiction. □

Lemma 5.

$0\notin I_4\left(G_3\right)$, and $1,2,3\in I_4\left(G_3\right)$.

Proof. 

In Figure 3, similarly, we have $b+f=2d$. Suppose $0\in I_4\left(G_3\right)$, then one of three edge labels of the top claw must be 1 or $-1$. Therefore, this implies $\pm 1=b+f=2d$ in ${\mathbb{Z}}_4$, and the solution for d does not exist, which is a contradiction. On the other hand, since $G_3$ is 3-regular by labeling 1, 2, and 3 over all edges, respectively, $I_4\left(G\right)\supseteq \{3,2,1\}\left(mod4\right)={\mathbb{Z}}_4^{*}$. □

Lemma 6.

For $k\ge 5$, $0\in I_k\left(G_3\right)$

Proof. 

See Figure 4.

□

Lemma 7.

For $x\in {\mathbb{Z}}_k^{*}$, $x\in I_k\left(G_3\right),k\ge 5$.

Proof. 

For the case of ${\mathbb{Z}}_6$, see Figure 5. For the general case $k\ge 5,k\ne 6,$ see Figure 6 and Figure 7, which give two ways of labeling to make the magic sum index x. It remains to show that at least one of them contains no edges labeled by zero. Do not suppose, say $x-2=0$ or $x+2=0$ in Figure 6, then in Figure 7, either $x+4=0$ or $x-4=0$ in ${\mathbb{Z}}_6$; however this has been ruled out already. □

To summarize from the above lemmas, we have the following result for the index sets of the graph $G_3$, namely Theorem 4:

$$I_k\left(G_3\right)=\left\{\begin{array}{c}\left\{0\right\},k=3,\\ {\mathbb{Z}}_4\setminus \left\{0\right\},k=4,\\ {\mathbb{Z}}_k,k\ge 5.\end{array}\right.$$

Now, we show that for 3-regular graphs, the sufficient condition admitting a perfect matching for having full index sets is also necessary. We start with the following lemma.

Lemma 8.

If G is a 3-regular graph such that $0\in I_4\left(G\right)$, then G admits a perfect matching.

Proof. 

Give G a ${\mathbb{Z}}_4$-magic labeling with vertex sum zero. Then, for each vertex $v\in G$, there must be an even number of edges incident to v with an odd label and thus an odd number of edges with an even label, which must be 2. However, it is not possible that all three edges are labeled 2, as then the sum of the edges incident to v would be 2. Thus, there must be one edge with the label 2 incident to each vertex. The set of such edges is the desired perfect matching. □

Theorem 3 follows directly from this lemma. We note that this theorem is not true for general regular graphs, in particular, it fails for 4-regular and 5-regular graphs, as shown in the following sections.

6. A 5-Regular Graph with Full Magic Sum Spectrum

In this section, an example is given for a 5-regular graph with a full magic sum spectrum, however, without any perfect matching. We start with the following lemma regarding 5-regular graphs:

Lemma 9.

Let $G=H\cup J$ be a 5-regular graph with a 3-factor H and a 2-factor J. Then, $I_k\left(G\right)={\mathbb{Z}}_k$ for $k\ge 4$, and $I_3\left(G\right)\supset \{1,2\}$.

Proof. 

If there is a solution to the equation $3a+2b=x$ in nonzero elements of ${\mathbb{Z}}_k$, then there is a labeling of index x in ${\mathbb{Z}}_k$ by labeling the 3-factor edges a and the 2-factor edges b. It is a classical result that the largest x with no non-negative integer solutions to $3a+2b=x$ is 1; thus, the largest one with no positive integer solutions is 6. We first look for positive integer solutions to $3a+2b=x$ for $x=k+3,\dots ,2k+2$. Note that a positive integer solution must have both a and b be less than k, or else $3a+2b$ will be at least $2k+3$. Thus, whenever $k\ge 4$, there is a solution to $3a+2b=x$ for $x=k+3,\dots ,2k+2$ with $a,b\in \{1,2,\dots ,k-1\}$; these correspond to nonzero solutions in ${\mathbb{Z}}_k$ to $3a+2b=x$ and in turn to ${\mathbb{Z}}_k$-magic labelings with index x. Additionally, when $k=3$, there is a solution for $x=2k+1=7$ and $2k+2=8$. This proves our lemma. □

Thus, to show that a 5-regular graph has full index sets for all k, it suffices to find a 2-factor and a ${\mathbb{Z}}_3$ magic labeling with index 0. Below, we give a 5-regular graph $G_5$ with a 2-factor marked in bold in Figure 8.

On the other hand, we also wish to show that $0\in I_3\left(G_5\right)$. In Figure 9, we label the bold edges 1 and the remaining edges 2.

This gives a ${\mathbb{Z}}_3$-magic labeling of index 0. Thus, G has full index sets for all $k\ge 3$.

We also observe the following Tutte condition for checking whether a graph contains a perfect matching:

Theorem 11

(Tutte, 1947 [21]). A graph G has a perfect matching if and only if, for every subset U of V, the number of connected components with an odd number of vertices in the subgraph with vertex set $V\setminus U$ is less than or equal to the number of vertices in U.

By Tutte’s condition in Theorem 11, selecting the four vertices in the square, this graph lacks a perfect matching. Thus, $G_5$ does not have a perfect matching, but it has full index sets for all $k\ge 3$. This resolves Theorem 5 for 5-regular graphs.

7. Magic Sum Spectra for Regular Graphs of Even Degree

We prove a series of lemmas to describe the index sets of even–regular graphs. Our knowledge of regular graphs of even degree again comes from the following theorem of Petersen [20]: A nonempty graph G admits a 2-factorization if and only if it is $2m$-regular for some $m\ge 1$. Note that T.-M. Wang and C.-M. Lin [22] observed that for any ${\mathbb{Z}}_k$-magic labeling f of a graph G with index r, one has the condition

$$2\sum _{e\in E\left(G\right)}f\left(e\right)\equiv r·\left|V\left(G\right)\right|\left(modk\right).$$

Therefore, we have the following result:

Theorem 12.

For a graph G with odd vertices and for even k, we have that $I_k\left(G\right)\subseteq 2{\mathbb{Z}}_k⊊{\mathbb{Z}}_k$.

This gives a limit to the maximum size of index sets of graphs with an odd number of vertices. We show that this bound is tight for even–regular graphs with an odd number of vertices.

Lemma 10.

Let G be a $2m$-regular graph with $m>1$, and let $k\ge 3$ be an odd integer. Then, $I_k\left(G\right)={\mathbb{Z}}_k$.

Proof. 

Pick some $x\in {\mathbb{Z}}_k$. Let $G=H_1\cup H_2\cup \dots \cup H_m$ be a 2-factorization of G. Arbitrarily select nonzero elements $a_1,\dots ,a_{m-2}$ of ${\mathbb{Z}}_k$. Then, select $a_{m-1}$ from ${\mathbb{Z}}_k-\left\{0\right\}$ such that $a_1+a_1+a_2+a_2+\dots +a_{m-1}+a_{m-1}\ne x$ (computed in ${\mathbb{Z}}_k$). Then, let A, taken as an integer in $\{0,1,\dots ,k-1\}$, be equal to the sum $a_1+a_1+\dots +a_{m-1}+a_{m-1}$ modulo k. Clearly, exactly one of $(x-A)$ and $k+(x-A)$ is even since k is odd; let $a_m$ equal half this even number. Now, we have that $a_1+a_1+\dots +a_m+a_m=x$ in ${\mathbb{Z}}_k$. Label each edge in $H_i$ with the label $a_i$; this gives the desired labeling and proves the theorem. □

Note that this theorem only relies on the fact that a $2m$-regular graph has two 2-factors; thus, this would hold for a regular graph of odd degree with two disjoint 2-factors.

Lemma 11.

Let G be a $2m$-regular graph with $m>1$, and let $k\ge 4$ be an even integer. Then, $I_k\left(G\right)\supset 2{\mathbb{Z}}_k$.

Proof. 

Pick some $x=2y\in 2{\mathbb{Z}}_k$. We use the same factorization as in the previous lemma. Label each edge in $H_1,\dots ,H_{m-1}$ with $\frac{k}{2}$ and each edge in $H_m$ with y. Then, the sum of the labels at each vertex will be $(m-1)k+2y=x$; this is the desired labeling. □

We note that our work is done for even–regular graphs with an odd number of vertices by Theorem 12, and we have now proved Theorem 6. On the other hand, for even–regular graphs with an even number of vertices, we continue as follows:

Lemma 12.

Let G be a connected graph, and let U be a subset of the vertices of order $2m$. Then, there is a subgraph of G consisting of m edge-disjoint paths, each of which has as its endpoint vertices in U such that every vertex in U is the endpoint of exactly one path.

Proof. 

We begin by choosing two arbitrary vertices in U and connecting them by a path. We then iterate the following process until all vertices in U have been used.

Select two vertices $u_1$ and $u_2$ in U which are not yet the endpoints of paths and connect them by a path $P_1$. Starting from $u_1$, follow the path to $u_2$ until reaching an edge that is shared by another path, should one exist. If $P_1$ shares edges with another path $P_2$, then eliminate these edges as follows: Assume that $P_2$ runs from $u_3$ to $u_4$. Furthermore, assume that the first edge on which they overlap, when walking from $u_1$ to $u_2$, is $e_1$ and that the last such edge is $e_2$. Assume that vertex $v_1$ is the endpoint of $e_1$ closer to $u_1$ along $P_1$ and that $v_2$ is the endpoint of $e_2$ closer to $u_2$ along $P_2$. Now, following $P_2$ from $u_3$ to $u_4$, assume that $v_1$ is reached before $v_2$ (this assumption is acceptable because $u_3$ and $u_4$ have not yet been distinguished).

Now, define the path $P_2^{\prime }$ from $u_1$ to $u_3$, formed by following $P_1$ from $u_1$ to $v_1$ and then $P_2$ from $v_1$ to $u_3$. This does not overlap any other paths since $P_1$’s first overlap is $e_1$ with $P_2$, and $P_2$ does not overlap any paths other than $P_1$. Next, define the path $P_1^{\prime }$ from $u_4$ to $u_2$, formed by following $P_2$ from $u_4$ to $v_2$ and then $P_1$ from $v_2$ to $u_2$. This can only overlap other paths between $v_2$ and $u_2$, and that section of the path has as many overlaps as $P_1$. Thus, by replacing $P_1$ with $P_1^{\prime }$ and $P_2$ with $P_2^{\prime }$, we reduce the total number of overlapping edges. We can continue doing this for the rest of $P_1$, and then $P_1$ will not overlap any other paths. Then, repeating this process for all pairs of points in U gives the desired subgraph. □

Lemma 13.

Let G be a connected graph with an even number of vertices. Then, G has a spanning subgraph with every vertex of odd degree. Call such a subgraph an odd-factor.

Proof. 

Let U be the entire set of vertices. We claim the graph described in Lemma 12 for this set of vertices satisfies this property. Denote this graph J and pick a vertex v. Then, in J, v is the endpoint of exactly one path. There may be other paths passing through v, but for each path with one edge incident to v, there must be a second such edge. Thus, v is incident to an odd number of edges in the path with v as an endpoint and an even number of edges from every other path, giving in total an odd number of edges incident to v. □

Lemma 14.

If G is a $2m$-regular graph with an even number of vertices, and $k\ge 6$ is an even integer, then $I_k\left(G\right)={\mathbb{Z}}_k$.

Proof. 

We first handle the case of $k=6,x=3$. First, we start with a ${\mathbb{Z}}_3$-magic labeling of the graph with index 0 and double each label. This gives a ${\mathbb{Z}}_6$-magic labeling of G with index zero, where each label is either 2 or 4. Now, let J be an odd-factor of G as in Lemma 13. For each edge in J, increase its label in G by 3. This will increase the sum of the labels at any vertex by 3 an odd number of times, so the induced sum will change by 3 and will thus be 3.

Now, we consider all other cases. First, choose a 2-factor H of G and label all remaining edges $\frac{k}{2}$. Let U be a set containing one vertex from each odd cycle of H. As in the previous proof, there are an even number of elements in this set. Thus, let J be the subgraph of G described in Lemma 12. We first label the edges of H with alternating numbers and then use J to fix the labels of U. If we have $x=1$, then set $a_{+}=\frac{k}{2}+1$ and $a_{-}=\frac{k}{2}$. If $x=3$, then set $a_{+}=\frac{k}{2}+2$ and $a_{-}=\frac{k}{2}+1$. Otherwise, set $a_{\pm }=\frac{x\pm 1}{2}$. Note first that $a_{+}+a_{-}=x$; additionally, note that the maximum possible value of either is $a_{+}=\frac{k}{2}+2$ in the case of $x=3$, which is at least two less than k in all cases except $k=6$. The minimum possible value of either is $a_{-}=2$ in the case of $x=5$. Thus, either value can be changed by $\pm 1$ without making it equal to zero. The label $\frac{k}{2}$ can also be changed by $\pm 1$ without making it equal to zero. For each path in J of even length, choose one endpoint and denote it a + vertex, and denote the other a − vertex. For each path in J of odd length, denote both vertices + vertices. Now, we label H. For each even cycle, label the edges alternating between $a_{+}$ and $a_{-}$. For each odd cycle, there will be a + vertex or a − vertex. If there is a + vertex, start from this vertex and label its incident edges with $a_{+}$, then continue to label this cycle alternating between $a_{+}$ and $a_{-}$; similarly for a − vertex, label its incident edges $a_{-}$. Finally, for each path in J, start with a + vertex and add $-1$ to the edge in J incident to it, then continue along this path alternating between adding $+1$ and $-1$ to the labels. This will only affect the vertex sums at the endpoints; for + vertices, it will decrease the sum by 1, while for − vertices, it will increase the sum by 1. For all other vertices, the sum will be $a_{+}+a_{-}+(\frac{k}{2}+\frac{k}{2})+\dots =x$; for − vertices, the sum will be $a_{-}+a_{-}+1+(\frac{k}{2}+\frac{k}{2})+\dots =x$; and for + vertices, the sum will be $a_{+}+a_{+}-1+(\frac{k}{2}+\frac{k}{2})+\dots =x$. Thus, this is a ${\mathbb{Z}}_k$-magic labeling with index x. □

Finally, we are in a position to complete the picture for all regular graphs of even degrees. We start with the following lemma for regular graphs of degree congruent to 2 modulo 4:

Lemma 15.

If G is a $2m$-regular graph (with m odd) and is connected with an even number of vertices, then $I_4\left(G\right)={\mathbb{Z}}_4$.

Proof. 

It suffices to show that either 1 or 3 is in $I_4\left(G\right)$; a labeling with index 1 can give a labeling with index 3 by negating each value, and vice versa. Let $H_1$ and $H_2$ be k-factors of G. Then, label each edge in $H_1$ label 3 and each in $H_2$ label 2. The sum at each vertex will then be congruent to $5k=k$ modulo 4. Thus, if $k\equiv 1(mod4)$, this is a labeling with index 1, and if $k\equiv 3(mod4)$, this is a labeling with index 3. This gives the desired labeling. □

We need the following classical fact to show the remaining cases:

Theorem 13

(Petersen, 1891 [20]). Let k be a positive integer. If a connected $2k$-regular graph G has an even number of vertices, then it may be k-factored. That is, G can be factored into the sum of two k-regular spanning subgraphs.

In the following, in particular, we show that a connected $4m$-regular graph of even order has a full ${\mathbb{Z}}_4$-index set $I_4$, and together with Lemmas 10 and 14, we resolve Theorem 8, i.e., any connected $4m$-regular graph G of even order has full index sets $I_k\left(G\right)$ for all $k\ge 3$.

Theorem 14.

If G is a connected $2m$-regular graph (with m even and $m\ge 2$) with an even number of vertices, then $I_4\left(G\right)={\mathbb{Z}}_{\mathbf{4}}$.

Proof. 

Let $k=2r$. By Theorem 9, we see G is 2-factorable. So, $G=R\oplus K$, where R is a 2-factor and K is a $(4r-2)$-factor. Duplicate edges for all $e\in E\left(G\right)$, then we obtain the new graph $G^{\prime }=G\oplus G=G\oplus K\oplus R=R^{\prime }\oplus R$, where $R^{\prime }=G\oplus K$ is a $(8r-2)$-factor. Note that $R^{\prime }=G\oplus K$ is connected as well, since G and $G^{\prime }$ are connected. Then, due to Theorem 13 by Petersen, the $(8r-2)$-factor $R^{\prime }$ can be factored into the sum of two $(4r-1)$-regular spanning subgraphs, say $T_1$ and $T_2$. Now, we define the edge labeling function $f^{\prime }$ on $E\left(G^{\prime }\right)$ as follows:

$$f^{\prime }\left(e\right)=\left\{\begin{array}{cc}-1\hfill & ,e\in T_1\hfill \\ 2\hfill & ,e\in T_2\hfill \\ -1\hfill & ,e\in R\hfill \end{array}\right.$$

Then, we see that $f^{\prime }\left(e\right)\in \{-2,4,-2\}$ and the vertex sum $g^{\prime }\left(v\right)=-(4r-1)+2(4r-1)-2=4r-3$ for all $v\in V\left(G^{\prime }\right)=V\left(G\right)$ over $G^{\prime }$.

Now, one may define a new labeling function f over G as follows:

$$f\left(e\right)=f^{\prime }\left(e\right)+f^{\prime }\left(e^{\prime }\right)+1,e\in E\left(G\right)$$

where $e^{\prime }$ is the duplication edge by e above. Therefore, the labeling function $f\left(e\right)\in \{-1,5,-1\}\equiv \{1,3\}\left(mod4\right)$ is feasible ${\mathbb{Z}}_4$ weights. Since G is $4r$-regular, over G one may have the vertex sum $g\left(v\right)=(4r-3)+4r=8r-3\equiv 1\left(mod4\right)$ for every $v\in V\left(G\right)$. Thus, $1\in I_4\left(G\right)$, and hence $3\in I_4\left(G\right)$ by symmetrically negating all edges since $-1=3$ in ${\mathbb{Z}}_4$. By Lemma 11 $\{0,2\}=2{\mathbb{Z}}_4\subseteq I_4\left(G\right)$; therefore, $I_4\left(G\right)={\mathbb{Z}}_4$. We are done. □

Therefore one can see from the lemmas and theorems above that we already resolved Theorems 6–8, which completely determines the index sets for all regular graphs of even degree.

Remark 6.

Note that Akbari et al. in [18] raised the following conjecture: Every connected $4k$-regular graph of even order admits a 1-sum 4-flow. As a byproduct, we already verified this conjecture, since $1\in I_4\left(G\right)$, as shown in Theorem 14.

8. Concluding Remark and Future Studies

Note that one may extend the above results to the case of the direct product of two Abelian groups, hence to the case of the finitely generated Abelian group, in terms of the observation below.

Lemma 16.

For any Abelian group $A=A_1\times A_2$, a graph G is A-magic with index $r=(r_1,r_2)$ if and only if there are two spanning subgraphs of G, say $G_1$ and $G_2$, such that $G=G_1\cup G_2$, $G_1$ is $A_1$-magic with index $r_1$ and $G_2$ is $A_2$-magic with index $r_2$.

Proof. 

Suppose f is an A-magic labeling of G, and $f\left(e_i\right)=(a_i,b_i)$ for $i=1,2,\cdots ,\left|E\right(G\left)\right|$. Let $G_1$ be the spanning subgraph of G such that $E\left(G_1\right)=\{e_j|f\left(e_j\right)\ne 0,\forall i=1,2,\cdots ,|E\left(G\right)\left|\right\}$ and define an $A_1$-magic labeling $f_1$ by $f_1\left(e_j\right)=f\left(a_j\right)$ for all $e_j\in E\left(G_1\right)$. Similarly $G_2$ is defined and so is $A_2$-magic labeling.

Conversely, suppose $G_1$ is $A_1$-magic and $G_2$ is $A_2$-magic under the labeling $f_1,f_2$, respectively. Define an A-magic labeling as follows:

$$f\left(e\right)=\left\{\begin{array}{cc}(f_1\left(e\right),f_2\left(e\right))\hfill & if e\in E\left(G_1\right) and e\in E\left(G_2\right),\hfill \\ (f_1\left(e\right),0)\hfill & if e\in E\left(G_1\right) and e\notin E\left(G_2\right),\hfill \\ (0,f_2\left(e\right))\hfill & if e\notin E\left(G_1\right) and e\in E\left(G_2\right).\hfill \end{array}\right.$$

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In this article, we study the properties of the magic sum spectrum for general graphs and determine the magic sum spectrum for complete bipartite graphs $K_{m,n}$. Also, we show that a regular graph G admitting a perfect matching has a full magic sum spectrum, namely, $I_k\left(G\right)={\mathbb{Z}}_k$ for all $k\ge 3$. An example is given for a 3-regular graph without any perfect matching whose magic sum spectrum is not full. Furthermore, we show that for 3-regular graphs, admitting a perfect matching is equivalent to having a full magic sum spectrum. Among other results, we completely determine the magic sum spectra for all regular graphs of even degree. The magic sum spectra for regular graphs of odd degree are not fully understood. However, for 5-regular graphs, we show that there exist examples without any perfect matching but with full magic sum spectra.

Here, we list the following open problems inspired by our work to be explored further:

• Characterize graphs whose magic sum spectrum is a subgroup of the Abelian group A, in particular for $A={\mathbb{Z}}_k$.
• Determine the ${\mathbb{Z}}_k$ index set for a 3-regular graph with no perfect matching in general.
• Find examples of r-regular graphs (r odd and $r\ge 7$) with magic sum spectra for all $k\ge 3$ but no perfect matching.
• Characterize r-regular graphs with magic sum spectra for odd $r\ge 5$.
• Characterize the graphs that have a full magic sum spectrum.
• Determine index sets of other nearly regular graphs, such as $W_n$, $F_n$, in addition to index sets of graphs without perfect matching in general.
• Generalize the constant-sum group flows to directed graphs and bidirected graphs.